A bucket containing water is revolved in a ve | vedclass

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(C) For the water not to fall from the bucket at the highest point of the vertical circle,the centripetal force must be at least equal to the gravitat

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$\frac{1}{2 \pi} \sqrt{\frac{g}{r}}$

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(C) For the water not to fall from the bucket at the highest point of the vertical circle,the centripetal force must be at least equal to the gravitational force acting on the water.At the highest point,the condition for the water to remain in the bucket is $m \omega^2 r \geq mg$.The minimum angular velocity $\omega$ is given by $m \omega^2 r = mg$,which simplifies to $\omega = \sqrt{\frac{g}{r}}$.Since the angular frequency $\omega$ is related to the frequency $f$ by the formula $\omega = 2 \pi f$,we can write $2 \pi f = \sqrt{\frac{g}{r}}$.Therefore,the minimum frequency of revolution is $f = \frac{1}{2 \pi} \sqrt{\frac{g}{r}}$.

$A$ bucket containing water is revolved in a vertical circle of radius $r$. To prevent the water from falling down,the minimum frequency of revolution required is ($g =$ acceleration due to gravity).

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